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Frequently Asked Questions on Chapter 4 – Quadratic Equations

Check whether the following are quadratic equations (x + 1)2 = 2(x – 3)?

(x + 1)2 = 2(x – 3) By using the formula for (a+b)2=a2+2ab+b2
x2 + 2x + 1 = 2x – 6 x2 + 7 Since the above equation is in the form of ax2 + bx + c = 0. Therefore, the given equation is a quadratic equation.

Quadratic Equations

Find the roots of the following quadratic equations by factorisation x2 – 3x – 10 = 0.

Given, x2 – 3x – 10 =0 Taking LHS,
x2 – 5x + 2x – 10
x(– 5) + 2(x – 5) (x – 5)(x + 2) The roots of this equation, x2 – 3x – 10 =0 are the values of x for which (x – 5)(x + 2) = 0 Therefore, x – 5 = 0 or x + 2 = 0
x = 5 or x = -2 No, not every chemical reaction is a redox reaction. Reactions like double decompositions, acid-base neutralisation reactions, precipitation reactions are non-redox reactions.

Quadratic Equations

Find the roots of the following quadratic equations, if they exist, by the method of completing the square 4x2 + 4√3x + 3 = 0.

4x2 + 4√3x + 3 = 0 Converting the equation into a2+2ab+bform, we get, (2x)2 + 2 × 2x × √3 + (√3)2 = 0 (2x + √3)2 = 0 (2x + √3) = 0 and (2x + √3) = 0 Therefore, either x = -√3/2 or x = -√3/2.

Quadratic Equations

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them 2x2 – 3x + 5 = 0.

Given,
2x2 – 3x + 5 = 0 Comparing the equation with ax2 + bx c = 0, we get
a = 2, b = -3 and c = 5 We know, Discriminant = b2 – 4ac
( – 3)2 – 4 (2) (5) = 9 – 40 = – 31 As you can see, b2 – 4ac < 0 Therefore, no real root is possible for the given equation, 2x2 – 3x + 5 = 0.

Quadratic Equations

Find the values of k for each of the following quadratic equations, so that they have two equal roots 2x2 + kx + 3 = 0.

2x2 + kx + 3 = 0 Comparing the given equation with ax2 + bx c = 0, we get,
a = 2, b = k and c = 3 As we know, Discriminant = b2 – 4ac
= (k)2 – 4(2) (3) = k2 – 24 For equal roots, we know, Discriminant = 0
k2 – 24 = 0
k2 = 24 k = ±√24 = ±2√6

Quadratic Equations

Check whether the equation (x + 1)^2 = 2(x – 3) is a quadratic equation

(x + 1)^2 = 2(x – 3)

By using the formula (a+b)^2 = a^2 + 2ab + b^2

x^2 + 2x + 1 = 2x – 6

x^2 + 7 = 0

Since the above equation is in the form of ax^2 + bx + c = 0.

Therefore, the given equation is a quadratic equation.

By factorisation method find the roots for the x^2 – 3x – 10 = 0 quadratic equation

x^2 – 3x – 10 = 0

Let us consider LHS,

x^2 – 5x + 2x – 10

x(x – 5) + 2(x – 5)

(x – 5) (x + 2)

x – 5 = 0 or x + 2 = 0

x = 5 or x = -2

The roots of the equation, x^2 – 3x – 10 = 0 are the values of x for which (x – 5) (x + 2) = Therefore, x = 5 or x = -2

Find the two sides if the altitude of a right triangle is 7 cm less than its base. And if the hypotenuse is 13 cm.

Given, the altitude of right triangle = (x – 7) cm

Let us consider, the base of the right triangle be ‘x’ cm.

From Pythagoras theorem, we know,

Base^2 + Altitude^2 = Hypotenuse^2

x^2 + (x – 7)^2 = 13^2

x^2 + x^2 + 49 – 14x = 169

2x^2 – 14x – 120 = 0

x^2 – 7x – 60 = 0

x^2 – 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

(x – 12) (x + 5) = 0

Thus, either x – 12 = 0 or x + 5 = 0,

x = 12 or x = – 5

Since sides cannot be negative, x = 12.

Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.

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